Functions — Page 13 | JEE Mathematics

Q121

For

x \in R - \left\{ {0,1} \right\}

, Let f1(x) =

1\over x

, f2 (x) = 1 – x and f3 (x) =

1 \over {1 - x}

be three given functions. If a function, J(x) satisfies (f2 o J o f1) (x) = f3 (x) then J(x) is equal to :

A f1 (x)

B

1 \over x

f3 (x)

C f2 (x)

D f3 (x)

Correct Answer

Option D

Solution

Given, f1(x) =

{1 \over x}

f2(x) = 1 $-$ x f3(x) =

{1 \over {1 - x}}

(f2 $\cdot$ J $\cdot$ f1) (x) = f3(x) $\Rightarrow$ f2 {J(f1(x))} = f3(x) $\Rightarrow$ f2{J (

{1 \over x}

)} =

{1 \over {1 - x}}

$\Rightarrow$ 1 $-$ J(

{1 \over x}

) =

{1 \over {1 - x}}

$\Rightarrow$ J(

{{1 \over x}}

) = 1 $-$

{{1 \over {1 - x}}}

$\Rightarrow$ J (

{{1 \over { x}}}

) =

{{ - x} \over {1 - x}}

=

{x \over {x - 1}}

Put x inplace of

{1 \over x}

$\therefore$ J(x) =

{{{1 \over x}} \over {{1 \over x} - 1}}

=

{1 \over {1 - x}} = {f_3}\left( x \right)

Q122

Let f(x) = x2 , x

\in

R. For any A

\subseteq

R, define g (A) = { x

\in

R : f(x)

\in

A}. If S = [0,4], then which one of the following statements is not true ?

A g(f(S))

\ne

S

B f(g(S)) = S

C f(g(S))

\ne

f(S)

D g(f(S)) = g(S)

Correct Answer

Option D

Solution

f(x) = x2 x

\in

R g(A) = {x

\in

R : f(x)

\in

A} S $\equiv$ [0, 4] g(S) = {x

\in

R : f(x)

\in

S} = {x

\in

R : 0 $\le$ x2 $\le$ 4} = {x

\in

R : –2 $\le$ x $\le$ 2} $\therefore$ g(S) $\ne$ S $\therefore$ f(g(S)) $\ne$ f(S) g(f(S)) = {x

\in

R : f(x)

\in

f(S)} = {x

\in

R : x2

\in

S2} = {x

\in

R : 0 $\le$ x2 $\le$ 16} = {x

\in

R : –4 $\le$ x $\le$ 4} $\therefore$ g(f(S)) $\ne$ g(S) $\therefore$ g(f(S)) = g(S) is incorrect

Q123

Let a function f : (0,

\infty

)

\to

(0,

\infty

) be defined by f(x) =

\left| {1 - {1 \over x}} \right|

. Then f is :

A not injective but it is surjective

B neiter injective nor surjective

C injective only

D both injective as well as surjective

Correct Answer

Option B

Solution

f\left( x \right) = \left| {1 - {1 \over x}} \right| = {{\left| {x - 1} \right|} \over x} = \left\{ \begin{array}{ll}{{{1 - x} \over x}} & {0 < x \le 1} \\ {{{x - 1} \over x}} & {x \ge 1} \end{array} \right.

$\Rightarrow$ f(x) is not injective but range of function is

\left[ {0,\infty } \right)

Remarks : If co-domain is

\left[ {0,\infty } \right)

, then f(x) will be surjective.

Q124

Let N be the set of natural numbers and two functions f and g be defined as f, g : N

\to

N such that f(n) =

\left\{ \begin{array}{ll}{{{n + 1} \over 2};} & {if\,\,n\,\,is\,\,odd} \\ {{n \over 2};} & {if\,\,n\,\,is\,\,even} \end{array} \,\, \right.

; and g(n) = n

-

(

-

1)n. Then fog is -

A neither one-one nor onto

B onto but not one-one

C both one-one and onto

D one-one but not onto

Correct Answer

Option B

Solution

f(x) =

\left\{ \begin{array}{ll}{{{n + 1} \over 2};} & {if\,\,n\,\,is\,\,odd} \\ {{n \over 2};} & {if\,\,n\,\,is\,\,even} \end{array} \,\, \right.

; g(x) = n $-$ ( $-$ 1)n

\left\{ \begin{array}{ll}{n + 1;\,\,\,\,n\,\,is\,\,odd} \\ {n - 1;\,\,\,\,n\,\,is\,\,even} \end{array} \right.

f(g(n)) =

\left\{ \begin{array}{ll}{{n \over 2};\,\,\,\,n\,\,is\,\,even} \\ {{{n + 1} \over 2};\,\,\,\,n\,\,is\,\,odd} \end{array} \right.

$\therefore$ many one but onto

Q125

Let A = {x

\in

R : x is not a positive integer}. Define a function

f

: A

\to

R as

f(x)

=

{{2x} \over {x - 1}}

, then

f

is :

A not injective

B neither injective nor surjective

C surjective but not injective

D injective but not surjective

Correct Answer

Option D

Solution

f(x) =

{{2x} \over {x - 1}}

f(x) = 2 +

{2 \over {x - 1}}

f'(x) = $-$

{2 \over {{{\left( {x - 1} \right)}^2}}}

< 0 $\forall$ x

\in

R Hence f(x) is strictly decreasing So, f(x) is one-one Range : Let y =

{{2x} \over {x - 1}}

xy $-$ y = 2x $\Rightarrow$ x(y $-$ 2) = y $\Rightarrow$ x =

{y \over {y - 2}}

given that x

\in

R : x is not a +ve integer $\therefore$

{y \over {y - 2}} \ne

N (N $\to$ Natural number) $\Rightarrow$ y $\ne$ Ny $-$ 2N $\Rightarrow$ y $\ne$

{{2N} \over {N - 1}}

So range $\notin$ R (in to function)